Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по алгебре 9 класс Макарычев Задание 384
Задание 384
\[\boxed{\text{384\ (384).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} 4x^{2} - 27x - 7 > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[4x^{2} - 27x - 7 = 0\]
\[D = 27^{2} + 4 \cdot 4 \cdot 7 = 841\]
\[x_{1,2} = \frac{27 \pm 29}{8} = - \frac{1}{4};7.\]
\[\left\{ \begin{matrix} 4 \cdot \left( x + \frac{1}{4} \right)(x - 7) > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in (7; + \infty).\]
\[\textbf{б)}\ \left\{ \begin{matrix} - 3x^{2} + 17x + 6 < 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} 3x^{2} - 17x - 6 > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[3x^{2} - 17x - 6 = 0\]
\[D = 289 + 4 \cdot 3 \cdot 6 = 361\]
\[x_{1,2} = \frac{17 \pm 19}{6} = - \frac{1}{3};6.\]
\[\left\{ \begin{matrix} 3 \cdot \left( x + \frac{1}{3} \right)(x - 6) > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in \left( - \infty;\ - \frac{1}{3} \right).\]
\[\textbf{в)}\ \left\{ \begin{matrix} x + 1 < 0\ \ \ \ \ \\ 2x^{2} - 18 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \\ x^{2} - 9 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 3)(x + 3) > 0 \\ \end{matrix} \right.\ \]
\[x \in ( - \infty;\ - 3).\]
\[\textbf{г)}\ \left\{ \begin{matrix} x - 4 > 0\ \ \ \ \ \ \ \ \\ 3x^{2} - 15x < 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x > 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x(x - 5) < 0 \\ \end{matrix}\ \right.\ \ \]
\[x \in (4;5).\]
\[\boxed{\text{384.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{2} + x^{2} - 6x + 9 - 9 = 0\]
\[2x^{2} - 6x = 0\]
\[2x(x - 3) = 0\]
\[x = 0;\ \ \ x = 3.\]
\[y = x - 3 = 0 - 3 = - 3;\]
\[y = x - 3 = 3 - 3 = 0.\]
\[Ответ:(0; - 3);(3;0).\]
\[x^{2} - 4x - 12 = 0\]
\[D_{1} = 4 + 12 = 16\]
\[x_{1} = 2 + 4 = 6;\ \ \ x_{2} = 2 - 4 =\]
\[= - 2.\]
\[y_{1} = x - 4 = 6 - 4 = 2;\]
\[y_{2} = x - 4 = - 2 - 4 = - 6.\]
\[Ответ:(6;2);( - 2; - 6).\]
\[x + 4x^{2} + 4x + 1 - 10 = 0\]
\[4x^{2} + 5x - 9 = 0\]
\[D = 25 + 144 = 169\]
\[x_{1} = \frac{- 5 + 13}{8} = 1;\ \ \ \]
\[x_{2} = \frac{- 5 - 13}{8} = - \frac{18}{8} = - 2\frac{1}{4} =\]
\[= - 2,25.\]
\[y_{1} = 2x + 1 = 2 + 1 = 3;\]
\[y_{2} = 2x + 1 = 2 \cdot ( - 2,25) + 1 =\]
\[= - 4,5 + 1 = - 3,5.\]
\[Ответ:(1;3);( - 2,25;\ - 3,5).\]
